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The Gershgorin circle theorem locates every eigenvalue of a square matrix using only its entries. A diagonal entry supplies a centre in the complex plane, and the sum of the off-diagonal absolute values in its row supplies a radius. Small off-diagonal couplings therefore confine the eigenvalues near the diagonal entries. This turns a global eigenvalue problem into a picture built one row at a time.

Precisely, let $A=(a_{ij})\in {\mathbb{C}}^{n\times n}$, with $n\ge 1$, and define the closed Gershgorin discsEvery eigenvalue $\lambda$ of $A$ satisfies $\lambda \in {\bigcup }_i{D}_i(A)$, including eigenvalues on the boundaries. No symmetry or diagonalizability is required. Despite the name, the theorem uses filled disks, not just their circular boundaries.^[ref]

The short reason is the triangle inequality. For a nonzero eigenvector $x$ with $Ax=\lambda x$, choose $i$ with $|x_i|={\max }_j|x_j|>0$. The $i$th row givesThus the largest eigenvector coordinate identifies a Gershgorin disc containing $\lambda$, even though that eigenvector need never be computed to use the bound.

This is useful when a full eigenvalue computation would be expensive: forming the Gershgorin discs needs one pass over the nonzero entries of a sparse matrix. If $0$ lies outside every Gershgorin disc, then $A$ is an invertible matrix, since zero is not an eigenvalue. The condition is exactly $|a_{ii}|>r_i$ in every row, namely strict diagonal dominance. This connects diagonal dominance to solvability of a linear system; the inverse infinity-norm bound from diagonal dominance adds a quantitative bound. It is a sufficient test: failure to exclude zero does not prove singularity.

For a concrete symmetric matrix, takeThe three Gershgorin discs have centres $2,5,9$ and are pairwise disjoint. The finite-dimensional spectral theorem makes the eigenvalues real, so their possible locations reduce to the intervals $[1.7,2.3]$, $[4.5,5.5]$, and $[8.6,9.4]$. Gershgorin eigenvalue counting gives exactly one eigenvalue in each. Independently, the characteristic polynomial iswhose numerical polynomial roots are $1.985811$, $4.989942$, and $9.024247$, rounded to six decimal places. All are positive, as Gershgorin bounds for symmetric matrices already guarantee without finding those polynomial roots.

S. Gerschgorin published the result in 1931 in Über die Abgrenzung der Eigenwerte einer Matrix, in the Bulletin de l'Académie des Sciences de l'URSS, issue 6, pages 749–754. The original paper's §2 states both eigenvalue inclusion and counting in separated groups of Gershgorin discs; §3 already discusses column bounds and transformations preserving the eigenvalues. Thus both the localization picture and ways to sharpen it were part of the original development.^[ref]

For the distinct-knot linear B-spline hats, let $h_i=t_{i+1}-t_i>0$. The mixed-normalization spline Gram matrix has entriesOnly indices in the finite basis are retained. This matrix has a row strict diagonal dominance margin at least $1/3$, so the inverse infinity-norm bound from diagonal dominance gives $\Vert G^{-1}{\Vert }_{{\ell }^{\infty }}\le 3$. Equidistant spline knots give off-diagonal entries $1/6$. Both pieces of every hat are retained; repeated endpoint spline knots require different boundary formulas.

For order six, the support-midpoint sites are $x_i=i+3$, soBy part (b), the complete finite matrix is the symmetric five-diagonal matrixOnly indices from $0$ through $n$ are retained; the first and last rows are truncated, with no exterior coefficients or wraparound terms. For a row with all possible neighbors, the strict diagonal dominance margin is $(66-26-26-1-1)/120=1/10$. Boundary rows have an equal or larger margin.

Here is the appropriate inverse estimate with its proof. If a square matrix $A$ has $|a_{ii}|-{\sum }_{j\ne i}|a_{ij}|\ge d>0$, choose an index $i$ maximizing $|z_i|$. The reverse triangle inequality givesThus $Az=0$ implies $z=0$, and a square finite matrix is invertible. Applying the same estimate with $z=A^{-1}b$ proves $\Vert A^{-1}b{\Vert }_{{\ell }^{\infty }}\le d^{-1}\Vert b{\Vert }_{{\ell }^{\infty }}$. The inverse infinity-norm bound from diagonal dominance consequently yieldsThis also proves existence and uniqueness of the requested interpolating spline for every data vector and every finite $n$. The quintic cardinal spline midpoint collocation matrix includes the cases of one or two coefficients, where the dominance margin is even larger.

The entries in (b) satisfyThus $G$ has strict diagonal dominance with margin at least $1/3$. To prove the inverse infinity-norm bound from diagonal dominance directly, choose $i$ with $|a_i|=\Vert a{\Vert }_{{\ell }^{\infty }}$. The reverse triangle inequality givesThis also proves that the kernel is zero and hence that $G$ is invertible. Substituting $a=G^{-1}b$ and taking the operator norm givesCombined with (a), it yields $\Vert P_S{\Vert }_{\infty }\le 3$ for linear splines, independently of the knot spacings. The diagonal-dominance argument already proves the required estimate.

Apply the general B-spline interpolation operator norm estimate in part (I) to the inverse norm from (b):and hence in particular $\Vert P_{\mathbf{x}}{\Vert }_{L^{\infty }}\le 3$, independently of $n$. The strict finite-dimensional estimate uses the actual finite collocation matrix; it does not require periodic closure, natural boundary conditions or endpoint knot repetitions.

There is also a direct shorter verification of the uniform bound. For an index $i$ maximizing $|z_i|$, the triangle inequality giveswith an even larger margin when a neighbor is missing. This is the inverse infinity-norm bound from diagonal dominance, and again gives $\Vert A_n^{-1}{\Vert }_{{\ell }^{\infty }}\le 3$.

The midpoint quadratic spline collocation matrix isIts strict diagonal dominance margin is at least $3/4-2/8=1/2$. The inverse infinity-norm bound from diagonal dominance thus already proves $\Vert A_x^{-1}{\Vert }_{\infty }\le 2$: choosing an index where a vector $c$ attains its maximum absolute entry gives $\Vert A_{x}c{\Vert }_{\infty }\ge \frac{1}{2}\Vert c{\Vert }_{\infty }$, which also proves invertibility.

To evaluate the finite-dimensional norm exactly, set $D_{ii}=(-1{)}^i$ and $B=D{A}_{x}D$. If $T$ is the matrix with ones on its two adjacent diagonals and zeros elsewhere, then $B=(3/4)(I-T/6)$. Since $\Vert T/6{\Vert }_{\infty }\le 1/3$, the Neumann serieshas nonnegative entries. Thus $A_x^{-1}=D{B}^{-1}D$ has absolute entry values equal to those of $B^{-1}$. Its absolute row-sum vector is $v=B^{-1}1$, characterized byThe homogeneous roots are $3\pm 2\sqrt{2}$, and the constant particular solution is $v_i=2$. Writing $q=3-2\sqrt{2}\in (0,1)$ and imposing the two boundary values givesThis is largest at the middle row or rows. Therefore, with $i_{\ast }=\lfloor (n+1)/2\rfloor$, the exact norm isFor example, it is $4/3$ for $n=1$ and $8/5$ for $n=2$, and it increases towards two as the system grows. Thus two is a uniform bound, rather than the exact finite-$n$ inverse norm.

A strict diagonal dominance argument proves the estimate even on a nonuniform strictly increasing spline knot sequence. Put $h_i=t_{i+1}-t_i>0$. Then $N_i$ has a rising piece of length $h_i$ and a falling piece of length $h_{i+1}$, while $M_i=2N_i/(h_i+h_{i+1})$. The same integrations as in part (b) give the linear-spline mixed Gram matrixOnly entries with indices in $1,\dots ,n$ are present. Thus ${\sum }_{j\ne i}|g_{ij}|\le 1/3$ in every row.

For $Ga=b$, choose an index $i$ with $|a_i|=\Vert a{\Vert }_{{\ell }^{\infty }}$. The reverse triangle inequality then yieldsHence $\Vert a{\Vert }_{{\ell }^{\infty }}\le 3\Vert b{\Vert }_{{\ell }^{\infty }}$. Applying this to every $b$ provesThis is the inverse infinity-norm bound from diagonal dominance; no total positivity theorem is needed. The proof also gives invertibility directly by taking $b=0$.

Sampling consecutive order-six Cardinal B-splines at their support midpoints gives the displayed finite B-spline collocation matrix. No exterior columns are retained. Its row strict diagonal dominance margin is at least $(66-2\cdot 26-2)/120=1/10$, so the inverse infinity-norm bound from diagonal dominance gives $\Vert A^{-1}{\Vert }_{{\ell }^{\infty }}\le 10$. The uniform knot samples are $N_{0,6}(0),\dots ,N_{0,6}(6)=(0,1,26,66,26,1,0)/120$, obtained recursively from the Cox-de Boor recurrence.